I'm trying to understanding the concept of invertibility in time series using the following example.
Consider the process $X_t = Z_t + \frac{1}{2}Z_{t-1}$, where $\{Z_t\}$ is a pure noise process with mean 0 and variance $\sigma_z^2$.
According to my text, this process is invertible since $f(x) = 1+\frac{1}{2}x$ has a single root, $x=-2$, which lies outside the unit circle.
So why is the process invertible? I think the reasoning is something like this: we can recursively solve for $Z_t:$ $$\begin{align*} Z_t &= X_t - \frac{1}{2}Z_{t-1} \\&= X_t-\frac{1}{2}(X_{t-1}-\frac{1}{2}Z_{t-2}) \\&= \cdots \\&= X_t - \frac{1}{2}X_{t-1} + \frac{1}{2^2}X_{t-2} - \frac{1}{2^3}X_{t-3} - \cdots \end{align*}$$
I see that the coefficients above are going to $0$, but how does this allow us to actually invert the process? What is the relationship between this observation and the fact that $f(x)$ above has all its roots outside the unit circle?